The Watchman's Walk Problem and its Variations
by
©
Rebecca K epingA thesi submitted to the School of Gra luate Studies
in partial fulfilment of the requir mcnts for the degree of
l\Iaster of cience
Department of Mathemati s and Statisti s Memorial Univ rsity of Newfoundland
N ovcmbcr 2009
t. John's Newfoundland
Abstract
Given a graph and a single watchman, the Watchman
's Walk Problem is concerned
with finding closed dominating walks of minimum length, which the watchmancan
traverse to efficiently guard the graph. When multiple guards are available, two nat- ural variations emerg :(
1)given a
fixed number of guards,
how can we minimizeth
length oftime for which vertices are unobserved? and(2) given fixed time constraints
on the monitoring of vertices, what is the minimum number of guards requir d? The present thesis reviews known results for the original problem as wellas its variations,
and proves an upp er bound on the number
of guards required when time is fix d.Acknowledgements
I would like to express a great deal of gratitude to my supervisors Dr. Danny Dyer and Dr. Chris Radford. The timely completion of this thesis would not have b en possible without Dr. Dyer's dedicated guidance. I would also like to thank my family and friends for their continued encouragement, and Johnathan for his support, patienc , and counsel.
Contents
Abstract
Acknowledgements
List of Figures
1 Introduction
1.1
Motivation .
1. 2 Definitions .1.3 Variations on the
problem
2 Fixed number of guards
2.1 One guard:
the original watchman problem . 2. 2 Mult
ipl guards:downsizing a dominating set
3 Fixed time
3.1
3.23.3
Introductory results .
.
..
. . . . .. A generalized upp
erbound for
odd tAnalysis of the bound
11
iii
Vl
1
1
4 69 9
19
27 27
33
40
4 Bounds for small even t 45
5 Conclusions and open questions 54
Bibliography 57
List of Figures
1.1 A minimum dominating set (shaded, left), closed dominating walk (centre), and MCDW (right) of a graph
G. . . . . . . . . . . . . . .
32.1 A closed dominating walk for
G
obtained from the spanning treeT.
12 2.2A
graph satisfyingw
1(G)
= 2IE(To)l that is not a tree. . . . . 13 2.3 Three choices for a spanning treeT ,
and the resulting dominating walks. 14 2.4 A graph of girth 4 for which w1 (G)<
2IE(T0 ) I for any spanning treeT.
15 2.5 A cactus G, the graphs G' and G" from Theorem 2.11, and a MCDW. 19 2.6 Different methods of monitoring a graph with two guards (g1 and g2). 213.1
A
graph G that is 2-monitored with four guards. . . . . . . . . 28 3.2 The length of time for which each vertex in G has been unobserved. 29 3.3 TI:·ees satisfying W1(T) = l n2
1j . . . .
3.4 A tree T with W3(T)
=
3> l n+z-
4j.
31 42 3.5
A
tre T satisfyingvVt
(T)= l n+z-
3j
fort odd and k=
t~3, 2 ::; j ::; k+ 1. 444.1 Possible subtrees Si when
t =
2 . . . . 4.2 Choices for the subtreeT'
when t=
4 (Case 1).4.3 Choices for the subtree
T'
when t=
4 (Case 2).48 49 52
Chapter 1
Introduction
1.1 Motivation
A museum
is attmpting
tomonitor its rooms. Each room is connec ted
to one ormore
otherrooms via hallway , and from
any givenroom it i
possible to see all adjacent room . Placing one guard in every room will enure all rooms arc constantly monitor d
, but thir quires more guards
than arenecessary:
weneed
only pla guards iu ~uch away
LhaL everyroom
ei.Lherh
as a guard ris adj
acent to ar
om with a guard. This problem belong
to the field of graph thery, and the set of room
werequir i · called
a dominating set.In graph theory, a graph G
is a et
V (G) of vertices together with a set E (G) of edges.Th
-dges
of G are subsets of size twofrom
V(G), and
we ay two vertices'U, v E V (G) arc adjacent or neighbou1·ing
if the edge {
u, v} (usually written uv)belong
toE(G). More generally, if
we allow formult iple edge
bctw en the samepair of vertice · t hen we
obtain a multigraph, andif we
a cept edges of the form uucalled loops, we obtain a reflexive graph. Here, however, the term graph will be restricted to what is sometimes called a simple graph: a graph with no multiple edges and no loops.
It is not difficult to see how the museum problem can be translated into the language of graph theory. The museum is a graph, say G, with rooms as vertices and halls as edges. The definition of a dominating set is then a set D ~ V(G) with the property that every vertex of G is either in D or adjacent to a vertex of D. The concept of graph domination is widely researched, and many results are known about the domination number of a graph: the size of a smallest dominating s t, d noted 1(
G) .
Minimizing the size of a dominating set is important, as 'wasteful' dominating sets ar a ·y to find (take D=
V (G), for example).A variation on domination, as introduced by Hartnell, Rall, and Whitehead in 1998 [4], consider an alt rnative method of guarding the museum: rather than placing one guard in each room of a dominating set, have a single guard (or 'watchman') walk around the museum in such a way that the visited rooms collectively form a dominating s t. This ensures that every room has been either visited by the guard or seen by the guard from an adjacent room. We will assume that the guard's route begins and ends in the same room, allowing the walk to be repeated.
In a graph, an alternating sequence of vertices and edges, such as the route of a guard through a museum, is called a walk; more formally, a walk of length k is a sequence v 0, e1, v1 , e 2 , ... , ek, vk where ei
=
vi-lvi for each i. Note that the length of a walk is the number of edges it contains. A walk is closed if it begins and ends on the same vertex and is dominating if the vertices of th walk form a dominating set.We see then that the desired route for a single museum guard as described above is
a closed dominating walk.
The
added economic efficiency of this method, as only one guard is r quir d for the wholemuseum
, is gained at the sacrificeof security, since at
any given time thr will b
e room that
are not visible by the guard.We will therefore be concerne l
again with minimality;in particular, we wa nt to find
a shortestroute for
the guard towalk. This
is theWatchman's Walk Problem:
given a graph G, find adominating walk
thatis closed
and of minimum length, or a minimum closed dominating walk (MCDW) in G.We will use
w1 (G)to denote the length
of aMCDW in
a graphG, wh re
the 1indicates
that a single guard iswalking G.
Although
a closed dominating walk can be construct d from a dominating set Dby forming
an alternating sequence of vertices and edges that at l astinclude
allvertices of D ,
thi isnot generally the most ffective method
, even ifD is minimum.
Figure
1.1illustrates this point;
infact
, from the graph G we see that aMCDW need not even contain a minimum dominating set. The watchman
's walk is thus a distinctly differ
ent problem from that of finding a minimum dominating set in a graph. MCDWs arefurther
xploredin
Chapter2. Before we introduce
the primary objective of thpresent thesis, a more thorough introduction to graph theory is required
; the following section provid s the
necessarybackground terminology.
G G G
Figure 1.1:
A minimum dominating set
(shaded, left), closeddominating walk
( cen- tre), and MCDW (right) of a graphG.
1.2 D e finition s
Th number of vertices IV(G)I
in a graph Gis called the
order of G, and the number of edges IE(
G)I
is calledthe
sizeof
G. Ifvertices
u and v are adjacent we say uis
a neighbourof
v, andthe
setof
allneighbours of
v along with v itselfis called
the closed neighbo'urhoodof
'U,denoted N[v].
If e=
'UV is the edge joining u and v then wesay
u and v areboth
incidentwith
e.The number
of edges incident with the vertex vin a
graph G is called the degreeof
v and isdenoted
degcv, or simply deg vif th
eassociated graph is clear from
context.A vertex of degree 0 is
called an isolate.A
graphwith
nvertices,
everytwo of which
are adjacent,is
call d the complete graphof order
n,denoted
Kn-A
bipartite graphis
onewhose v r
tices can be palti-tioned
intotwo
setsA
andB
suchthat
every edge in the graph hasone end in A
and the other in B; similarly, a multipartite gmphhas its vertex
setpartitioned into multiple sets such that no vertex has a
neighbour in its own set. The term 'complet ' is applied to abipartite or multipartit
e graph when allpossible
dges arepresent. A
complete bipartite graphthat has one set of size 1 and the others
tof size k is called
a k-star.The concept of a
walk in a graph, as introduced in Section 1.1
, leads to anumber
of
furtherdefinitions. For
example, a u-v walk is awalk beginning
on the vertex u and endingon the vertex
v.A closed walk with
all edgesdistinct
is called a circuit, and a walk with both vertices and edges distinct is called a path.A closed
circuit withno
repeatedvertices
exceptfor the first
andlast is
called a cycle; a cycle of length k is called ak-cycle
andis
d notedCk.
If a circuitin a graph
G visits every edge of G exactlyonce then
itis
calledan
Eulerian circuit, andwhen such
awalk
exists G- - - -- - - -
is said to be Eulerian. It is a well-known result, originally observed by Euler, that a graph is Eul rian if and only if each of its vertices has even degree. A Hamilton cycle in a graph G is a cycle which includes every vertex of G exactly once, and if such a cycle exists then th graph G is said to be Hamiltonian.
A graph His a subgraph of a graph G if V(H) ~
V(G)
and E(H) ~ E(G). A spanning subgraph of G is a subgraph of G with vertex set V (G). An induced subgraph of G is a subgraph H whose edge set E(H) consists of all edges of G that have both endpoints in V (H). For a set of vertices S ~ V (G) we use G \ S (or G \ v if S contains a single vertexv)
to denot the induced sub graph with vertex setV
(G) \S,
and for a set of edges S ~ E(G) we denote by G \ S the subgraph with vertex set V(G) and edge set E(G) \ S.A graph is said to be connected if there is a path between any two vertices, and the maximal connected subgraphs of a disconnect d graph are the components of the graph. If the graph G is connected and the graph G \ v is disconnected then the vertex v E V (G) is called a cut vertex; similarly, an edge whose removal disconnects the graph is called a cut edge. A maximal connected subgraph containing no cut vertices is called a block of the underlying graph. A cactus i a graph with the prop rty that each of its blocks is either an edge or a cycle.
The girth of a graph is the length of a shortest cycle in the graph. The girth of a graph that contains no cycles is defined to be infinity. An important family of graphs called trees are categorized by the abs nee of cycles; equivalently, a tree is a graph which has a unique u-v path for any two vertices u, v. ote that this definition forces trees to be connected. A spanning tr·ee of a graph G is a spanning subgraph of G that is a tree. Vve refer to vertices of degree 1 in a tree as leaves and to a leaf's single
neighbouring vertex as a stem.
If T is a treethen
L(T)denotes the set of leaves of
T, and wewill define T
0to be t
he leaf-deleted subtree T \ L(T).If a graph G has a u-v path
then t
he distance in G from u to v,written
dc(u, v)(or
d(u, v)when G
isclear)
, isthe
lengthof a shor test u-v path in
G. If Sis
aset of vertices
inG then the distance from
avertex
v (j.S to
theset S
is givenby
d(v, S) = min d(u, v).uES
We
move
now from basic background terminology to a few sp
ecific concepts thatwill appear in forthcoming discussions: matchings and pair
eddominat ion. A matching
ina gr
aph G isa
setof edges of G
that hav no
common endpoints.A
maximum matching is one containingthe greatest number of edges, and
a perfect matching isone which
uses every vertexof the
graph. A total dominating set of a graph G is adominating set of G
with theproperty that
everyvertex
of Gha · a neighbour
in D.At first this may
not appear tobe different from the original definition
of a dominating set; however, the setof shaded vertices in F
igure1.1
is an exampl
of adominating set t
hat is not a total dominating set, sinceneither of the shaded
verticeshas a
neighbourin t
hedominating set
.Finally, a total dominating
set D is called a paired dominating set if thesubgraph induced by D has a perfect matching.
1.3 Variations on the proble m
Two variations of the original watchman's walk problem ar
e considered in the present
thesis.
Both are motivated by supposing that multiple guards are availab
le to monitora network.
vVhen determining routes formult iple guards on a single graph, a balance
is sought between security and economy:we want to minimize b
oth the t
ime for whichvertices
areunobserved
aswell
asthe numb
er of guardswe must hire, but the two
arenegatively correlated
. Inthe second half of
Chapt
er 2 we summarize the resultsof Hartnell
and Whitehead's Downsizing a dominating set [6],where t he priority is
given to economy -they
assume afixed number of guards and
attempt tomonitor
the graph as efficiently aspossible wit h
those guards.In
Chapter 3 we consider
the opposite problem, expanding on a variation first introduced by Davies, Finbow,
Hartnell,Li
and Schmeisser in[1]. Here
we assumthat a museum cares less about
how many guards are employed than aboutprotecting
itsvaluables . The museum may requir
e, for example, that each room must be
seen every 10 minutes. The goal is to respect
this t
ime r straint whileusing as few guards
aspossible.
We will
say avertex is
unobserved if neither the vertex nor
any of its neighbours isoccupi
ed by a guard.Hence for
a given graph G and length of time t,
we areinterested in findin
gthe minimum numb
er of guardsneed
ed todominate the graph
such that no vertex
isunobserved for more t han
t consecutive units of
time. Moreformally, for fixed time
t EN ,
a graph G can be t-monitored by a set S of guardsif there exist
s afunction f : S
x N ----.V
(G) such that
(i)
For
every guard g E Sand at every timeT E N, f(g, T+ 1)
E N[f (g,r)],
and(ii)
For
everyvertex v
E V( G) and every interval I C N of length t+
1, there exists
aguard
g ES
and a timeT E I such that f(g,r )
EN[v].
Not
e thatf
(g,r)
isthe ver
tex occupied bythe guard g
at time r . Essentially, condi-
tion (i)
ensures that at each unit of time, guards may onlymove fr
om avertex t
o oneof its neighbours (i.
e., no 'jumping' is allowed), and condition
(ii) ensures that everyvertex has a guard within
its closed n ighbourhood at least once every t+ 1 units of time. For
a given graphG
and length of timet
, denote byvllt(G)
the minimum valuof IS I ,
the number of guardsneeded
tot- monitor a
graph.In [1], the authors
find
upperbounds on Wt
(T) for t :::;3 when
T is atree. The
primary obj
ective of the present thesis isto
generalizethe results
of[1] by
finding an upper bound on Wt(T) fort> 3. An upper bound that
holds for all oddnatural
numbers
t ispresented in Chapter 3. This is followed
by an analysis of the bound, including ad
escription of afamily of trees for
which itis
attain d.In Chapter
4 weprove bounds for
small even values oft(t =
2 andt = 4). Finally, in
Chapter 5 wediscuss a conjectured
uppr bound for
all even values of t and suggest otherfu
turedirections for this research.
Chapter 2
Fixed number of guards
In this chapter we review the original watchman's walk problem as well as the 'down-
sizing'
variation. Both of these problemsconsider
optimal methods of monitoringa graph given a
fix d number of guards. 'vVe begin with a singlguard, the results for which are
primarily from [4].2.1 One guard: the original watchman proble m
Recall that w1 (G) is the length
of a
minimumclosed
dominatingwalk (M CDW)
ina connected graph G.
Questions ofcomplexity are among the
firstconsidered
forgr aph
theory problems like the watchman's walk; Hartnell, Ralland
Whitehead[4]
show
that findinga
MCDW is NP-complete forgeneral graphs.
The proof involves relating the watchman's walk problem tothe
well-known Hamilton cycle problem:given a graph
G, does thereexist a
Hamilton cycle in G? This problem is famously NP-complete [3],and
we willsee
how it can be usedto show
the same is trueof the
watchman's walk problem. Let CLOSED DoMINATING WALK be phrasedas
follows:. - - - -- - - - --
given a graph G and positive integer k, is w1 (G)
<
k? Then we have th following result.Theorem 2.1.
[4]
CLOSED DOMINATING WALK is NP-complete.Proof. Note firstly that CLOSED DOMINATING WALK is in NP, since it is straightfor-
ward to verify any solution to the problem. Given a graph
G
of order n, takek =
n in the decision problem and create a new graphG'
by attaching a degree-one vertex to every vertex of G. A MCDW in G' need not visit any of these degree-one vertices, but must visit their neighbours in order to monitor all vertices. Thus every vertex inG
must be included in a MCDW ofG';
we can conclude that w1 (G') :::::: k ,
since there arek
vertices inG.
IfG
is a Hamiltonian graph then there exists a closed walk of lengthk
containing every vertex of the graph, and in this casew
1 (G') = k.
H nc if w1(G')> k
thenG
is not Hamiltonian, and if we could find a MCDW inG'
of length k then w could find a Hamilton cycle in the arbitrary graph G, a problem we knowto be NP-complete. 0
We will see that despite the level of complexity for general graphs, there are many types of graphs for which the watchman's walk problem is very approachable. Indeed, the following two lemmas will completely solve the problem for trees.
Lemma 2.2.
[4]
Every cut vertex of a graph G must belong to every dominating walk of G.Proof. Let v be a cut vertex of G and let W be any dominating walk of G. If HI is the trivial walk on the single vertex v then we are done; otherwise let G1 be a component of
G \
v that contains a vertex ofvV
and let G2 be a second component ofG \ v. If HI
does not pass through
vthen it does not reach vertices of G
2,as
vis t he only vertex in
Gconnecting those components.
If uis a vertex in G
2then
uis not on
Wand consequently must b e adj acent to a vertex on the walk. So
uis adjacent to a vertex of G
1 ;but then G
1and G
2are not separate components of
G \v, which is a contradiction. Hence every cut vertex belongs to every dominating walk of
G,as
claimed. D
Lemma 2.3.
[4]
Let G be a connected graph of order at least 3. If W is a MCDW in G then HI does not include any vertices of degree 1.Proof.
To reach a vertex v of degree 1 the walk must first visit the single neighb our of
v,from which
itcan dominate
v;it is therefore unnecessary to add the two extra
edges required to visit v
itself.D
Not
inparticular that a MCDW in a tree does not include any
leaves. As sug-gest ed
,the two preceding
lemmastell us exactly how to find a MCDW for any tree.
Since every non-leaf vertex of a tree is a cut vertex, we know the vertex set of any MCDW in a tree will include all non-leaves a nd no leaves; i. e., the vertex set is always
V(T) \ L(T),for a tree
T.Since a MCDW must return to t he vertex it starts on, and since t
here is only one pathbetween any two vertices of a t ree, it
is easyto see that every edge traversed by a closed walk will in fact b e traversed twice when the graph
isa t ree. Recall t
hat T0is the tree
T \ L(T);then we
have shown w1(T):=:: 2IE(To)l.
Let us find a dominating walk in T.
Ifwe double every edge of T
0t hen every vertex
has even degree and hence there exists an Eulerian circuit in this new tree. Since t he
vertices traversed are all t
henon-leaves ofT , this circuit
is a closed dominatingwalk
ofTof length 2IE(To)l. A MCDW will be at most this
length, so w
1(T):::; 2IE(To)l.
We thus have th following theorem.
Theorem 2.4.
[4]
1fT is a tree then w1(T)= 2IE(To)l,
and an Eulerian circuit in the tree T0 with doubled edges is a MCDW forT.Theorem 2.5.
[4]
For a connected graph G and any spanning tree T of G,w
1(G) ::;
2IE(To)l.
Proof. Let T be any spanning tree of the graph G. We know that a MCDW for T has length
2IE (To)l.
But sinceV(G)
=V (T) ,
this walk is also a closed dominating walk of G, and it follows that a minimum closed dominating walk of G has lengthat
most
2IE(To)l.
DFigure 2.1 illustrat
es
the method describedabove
for finding an upper
bound onw
1(G)
. otet
hat the walk obtained is nota
MCDW,since
traversing one of the 6-cycles (e.g., the shaded vertices) int
his graph gives a shorter closed dominating walk; however, we can
at least conclude thatw
1(G) ::;
10.o :
I I I
0 0 0
Figure 2.1: A closed dominating walk for
G
obtained from the spanning treeT.
We have already
established
that treesattain the
upper boun l of Theorem 2.5, sinceevery non-leaf edge
is traversed twice in a MCDW fora
tree. Figure2.2 shows
, - - - - -- - - -
a graph
that is
not a tree that also satisfies w1(G) 2IE(To)l,
for the indicated spanning treeT.
G ZT0 , ,
0 0
'o=--= --==-=o
'
'0 (j
Figure 2.2: A
graph satisfyingw
1 (G)= 2IE(T
0 )I
that is not a tree.An important
note here is that for a given graph G,different choices for
the span-ning
tree Twill
likely resultin
differentvalues for
IE(To) 1.Specifically,
a spanning tree with many leaveswill
resultin T
0having fewer
edges.Consider again the graph
G in Figure2.2. Figure 2.3
showsthree
different spanning trees of G and the corre- spondingclosed dominating walks of G
for each. We see that theupper bound given in Theorem 2.5
canbe
slightly improved ifwe
specifythat
the spanning treeT
be the 'best' spanning tree; i.e.,that we
choose T sothat
To has thef west number of
edges. This is equivalent to finding a spanning tree of G with themaximum number
of leaves, aproblem which
isknown to b
P-hard [3].The
followingtheorem categorizes a class of graphs that do
notme
tthe bound of Theorem 2.5 for
any choiceof spanning tree
.Theorem 2.6.
[4]
Let G be a connected graph and letT be any spanning tree of G.IJG
has girth at least7
then w1(G) < 2IE(To)l.
Proof. Assume
a
graph G has girth at least7 but that w
1(G) =2IE(To)l for
some spanning treeT of G.
Let u and vbe the
endvertices
of some edgein G
that isnot
inT .
LetP be the
unique u-vpath
inT ,
andlet
u' and v'be the neighbours
. - - - -- - - - -- - - - · - - - -
T T
Figure 2.3: Thr
choices for a spanning tree T , and
the resulting dominating walks.of 'U and v, resp
ectively, on P.
Since every vertex onP
has degree at least two inT
(except possibly ua
nd v), this path is contained inTo .
In particular, u' and v' are inTo.
Note that dr0 ( u', v')>
3, since otherwise sucha
path from u' to v' together with { u', u'u, u, uv, v, vv',v'}
would form acycle
of length 6 in G, which contradicts that the girth of G is at least 7.Now, double each edge of T0 and let vll b
e an the
Eulerian circuit in the resulting multigraph. Thiscircuit has length
2IE(To)land
is thus a MCDW ofT. But if we replace one occurrence of the edges of P ~ E(T0 ) from u' to v' (of which there areat
least 4) onv\l
with the edges u'u, uv, vv', then we obtain a walkinG that is at least one edge shorter than lill. This new walk has all vertices of Wand so
is dominating, which contradicts the fact that W is a MCDW. Thus, there is no suchspanning tree
of G; that is, no
spanning tree T satisfies w1 (G)=
2IE(T0) I,as required.
0The girth requirement cannot be tightened here, as there do
exist graphs of girt
hsix
that attain the bound in Theorem 2.5 (a cycle of length six, forexample)
. Also note that the converse of Theorem 2.6 is not true, as Figure 2.4 demonstrates that theupper bound is not
attained for every graph of girth less than
seven.We can
see thegraph
Ghas w
1 (G)< 2jE (T
0 )I for
every spanningtree
Tbecause up
to isomorphismthere
is only one suchT, with the corresponding walk having length 6
, and traversingthe 4-cycle in
G gives a shorter closed dominatingwalk.
G
0 D
Figure 2.4: A
graph of girth 4 for which w1(G) < 2jE(To )l
for any spanning tree T .
The fo
llowing theorems consider the watchman's walk problem for several ammontypes of graphs.
Theorem 2. 7.
[4]
If G is a connected graph then w1 (G) =0
if and only if G has a dominating verte.'E (that is, a dominating set of size 1).Proof.
This is trivi
al; the watchman need notmove from the dominating v rtex.
DTheorem 2.8. [
4]
Let G be a complete multipartite graph. If any part is a single vertex then w1(G)
= 0, and otherwise w1(G)
= 2.Proof. If
one part of a complete multipartite graph is a single vertex, then
that vertexdominat
esthe entire graph and so, by
Theorem 2.7, w1(G) =0. Ot
herwise, a vertex udomin
atesthe v rtices
in allother p
arts except its own, which can be dominatedby one vertex, say
v, from any other part. Since G is complet
e, u and v are adjacent and theclosed walk of length 2 between
them is aMCDW. Thus w
1(G) = 2 in t
hiscase. D
,---~---- - - - · - - - -
Theorem 2.9.
[4]
Let G be a connected bipartite graph with bipartition (A, B), where bothA
andB
contain at least2
vertices. LetA'
denote a minimum subset ofA
that dominates all ofB ,
and letB'
denote a minimum subset ofB
that dominates all of A. Thenw 1(G)
~ 2(ma.~{IA'I,IB' I} ).
Proof. Since G is bipartit , no vertex of A dominates any other vertex of A. Likewise for
B.
Hence, ifA"
is the subset of vertices fromA
on a MCDW thenA"
must dominate B and consequently has at leastlA' I
vertices. Similarly, the set of verticesB"
fromB on
a MCDW must hav size greater than or equal toIB'I·
Since we must enter and leave each vertex of the larger of the two setsA "
andB ",
our MCDW has length at least twice the cardinality of the larger set, which is at least the larger ofA'
andB' .
0Theorem 2.10.
[4]
If Cn is a cycle of length n thenifn ~ 6 if 3 ::::; n
<
6Proof. If G is a cycle then we have two clear choices for a 'good' closed dominating walk; either we walk the entire way around the cycle, making a walk of length n, or we
walk partially around the cycle in one direction befor r v rsing and returning to the starting vertex. With any other walk there will be edges traversed more than twice, which adds unnecessary length. Label the vertices of Cn as v1, v2, ... , Vn· Beginning at v3 and walking to Vn ensures every vertex is observed, since the guard can see v1 from Vn and v2 from
v
3 . Reversing direction at Vn and returning to 'V3 creates a closed walk that is minimal in the sense that if we had reversed at any vertex beforVn then the walk would not be dominating (v1 would be unobserved). This method
.---~~- -
gives a walk of length 2(n ~ 3), which will be shorter than a complete traversal of the cycle if 2(n ~ 3)
<
n, or n<
6. Hence, w1(Cn) =
2(n ~ 3) for n:::; 6, and otherwisewl(Cn) = n .
DNotice that in the proof of Theorem 2.10 we are given, in addition to
w
1(Cn),
precise constructions for MCDWs in n-cycles. Results about w1(G)
and MCD\,Y constructions are known for other families of graphsG.
Given two graphsG
andH ,
the Cartesian product graphGDH
is the graph with vertex setV(G )
xV(H)
and edge set{(u,v)(u ,v')iu
EV(G) , vv'
EE(H)}
U{ (u ,v)(u'v) lv
EV(H) , uu'
EE(G)}.
In [5], forT
a tree, sharp bounds are found for w1(TDKn) ,
and nee ssary and sufficient conditions are found for a walk in TDK2 to be a MCDW. In [4], the following theorem describes MCDWs in cactus graphs.Theorem 2.11.
[4]
Let G be a connected cactus. Let G' be the induced subgraph of G obtained by deleting all vertices of degree 1, all vertices of degree 2 that are on 3- cycles, and exactly one pair of adjacent vertices of degree 2 from each cycle of length4
or 5 that contains such a pair of vertices. If each wt edge of G' is duplicated to formG",
thenG"
is Eulerian and any Eulerian cirwit inG"
is a MCDW ofG.
Proof. Let G be a cactus graph. W will show that an Eulerian circuit formed as described above is a MCDW by showing that none of the identified vertices need to be on a dominating walk, that each of the remaining vertices (those in G') must be on a dominating walk, and that every edge of
G'
must be traversed in order to connect its v rtices. Doubling th cut edges follows necessarily to ensure that the walk is closed.By Lemma 2.3, no vertex of degree 1 needs to be on a MCDW, so we eli card
such vertices (i. e., we do not
include them in G' ). Any vertex of degree 3 or higher
,or of d egr 2 and not on a cycle,
is a cutvertex in a cactus. To see
this, note thatsince every b
lock is a cycleor an edge, a ver tex belongs t o single block if and only if it
ison a cycle a nd
hasdegree two; otherwise the ver tex belongs to two blocks and its
removalwould disconnect t hose blocks. Cut vert ices must be on
anydominating walk, by Lemma 2.2, so we keep these vertices in G'.
V ert
icesof degree 2 on a 3-cycle will be seen fr om the cut vertex (or ve rt ices) on th e cycle, so discard them.
Ift here are adjacent ver tices of degree 2 on a 4-cycle then we discard one pair of them and keep the two remaining a dj acent vert ices, from which a guard can monit or the discarded p air. The edge between t he retained ver tices must be on a dominating walk
inorder for the guar d to m ove from one v r t x to t he other.
Ift
here are no adjacent vertices of degree 2 on a 4-cycle t hen there are vert ices of degree 3 or h
igher(i.e., cut vertices) at opp osite corners, which must be on
adomin at
ing walk. Fora guard t o move between
these opposite ver
tices,two adjacent edges must b e traversed , and if the guard 's walk is t o b e closed the n two dg s will hav to b e traversed
in theopposite direct
ionas well; we can t herefore put all four vertices and all four edges of the cycle in G' . A similar rule applies for 5-cycles.
For cycles of
length 6or more, a complete tr aver sal suffices for th minimum domin ating walk, even if that t raversal is interrupt ed (at cut vert
ices), and soG' will
includefull cycles of any
length higher than5.
Now, any vertex
inG'
is either a cut vertex or is on a cycle and has degree 2, an l
each cut vertex
is either the endof a cut edge or only belongs t o cycles.
If avertex
only b elongs to cycles then each of it s cycles contr
ibutes2 incident edges and so t he
total degree of the vertex
is even. Thus,if we duplicat e each cut edge of G' t o create
G",
all
verticesof the result
ing graph have even degree. We see that to dominate thecycles
of G wecan walk each edge
of G',and
toconnect these
cycl s witha
closed walkeach cut edge must be walked twice. Such a walk
is obtained pr cisely by findingan Eulerian circuit in G".
0Figure 2.5 below demonstrates how Theorem 2.11
applies to t
he given cactus graph G.G G'
Figure 2.5: A
cactus
G, the graphs G' and G" from Theorem 2.11, and a MCDW.2.2 Multiple guards : downsizing a dominating set
The original watchman's walk probl m
can
be viewedas an attempt to minimize t
he unobserved time of vertices, givena single guard.
In [6]t
his problem is generalizd to
multiple guards, but the question is stillessentially
the same:given
a fixednumber
of guards, how can we minimize the length of time for which vertices are unobserv
d?
The precise problem addressed in
[6]
i motivated as follows.Suppose firstly that
a
museum or other network has enoughguards to
place oneat each vertex of a dominating s
t, so thatall ver tices ar e under constant
monitoring.Let D b
e a
dominating set. If we hav IDIguards a
ndach remains st ationary at a
vertex of D, then we have anextr mely efficient but expensive security
network.Now sur pos t
hat the guards hav- b- n lownsized,so that only some
fraction q,0
<
q<
1, of the guards are now employ d. The following questi n aris s
:given
qjDIguards, how can we minimize th maximum time for which any v r tex is unobserv
d?Given
a closed d ominating walk
in a graph and mult
iple guardat our disposal, a
natural strategy is to have
the guards 'share' t
he dominating
walk, by spacing them out along it a· equally as possible. This will notalways be
the most effective method,as illustrated in Figure 2.6:
if two guards share the closeddominating walk on t
he left, which has length 12, then the leaves of this tree are unobserved for5 cons cutive units
of time, whereas with the two disjoint walks on the right no vert x i unoberved for
more than 3 units of time. The inefficiency is even more marked when we note that the closed dominating walk in this cas is actually minimum. However, the methodof sharing a
dominating walkat least gives
us an upper bound on the length oftime
for which vertices must be unobserved.
Lemma 2.12 formalizes this
ida, which i
used repeatedly
in [ 6].Lemma 2.12. If a graph has a closed dominating walk of l ngth m then it can be dominated with p guards such that no veTtex is unobserved fo1· more than
l ~ l -
1units of time.
Figure 2.6:
Different methodsof monitoring a grap
h with two guards (g1 and g2 ).Proof. If p guards are spaced out as evenly as
possible along
aclosed walk
of lengthm
,then any two
guardswill be
atmost I~ l
edges apart. If the guardsfollow
one another along the walk then everyvertex
on thewalk is occupied
at least once veryI ~ l units of time. Since
thewalk is dominating, this means
every vertex in the graph is observed (perhaps from
aneighbour)
at least once everyI~ l
units of time,or
equivalently no vert x is unobserved for more thanI~ l -
1units of time.
DThe bound
given in Lemma 2.12would
obviouslybe str
engthened ifthe
closeddominating walk was of minimum
length,but
sincefinding a MCDW in
a general graph is computationally difficult,we
settlefor
acleverly
constructed closed walkwhose vertices
containa
givendominating
set D.This
construction is outlinedin Theorem 2.14; first
we needthe following
lemma.Lemma 2.13. If D is a dominating set in a connected graph G then for any set of vertices S <;;;; D there exists a vertex v E D \ S such that de ( v, S) :::; 3.
Proof.
Suppose there
exists a subset Sof
D for which every vertex vin
D \ Shas
dc(v , S)
~4. Let v be
anyvertex
inD but not in
Sand chooseu to b
e the closestvertex in S to v.
LetP = u ,
v1, v2 ,v
3, v4, . . . ,v
be a shortestu- v
path inG. The
vertex
v2 isnot in S nor
adjacent to anyvertex in S because
oth rwise u is not the
close t vertex to v
inS. Furthermore, v2 and its neighbours are not in D \ S, because- - - -
these vertic
sare within distance 3 of
u E S and by assumption the set Sis
at leastdistance 4 from
anyvertex
in D \ S.But then
v2 is not in D and is not adjacent toa vertex of D , which contradicts t
he fact that Dis adominating
set inG.
D Theorem 2.14.[6]
If G is a connected graph with dominating set D then G can be monitor-ed withqJDI
guards,0 <
q<
1, such that no vertex is unobserved for- more thanj% l -
1 units of time.Proof. Let v
be
anyvertex
in D.Construct a
subtree T of G containing th vertices
of Dvia t
he following iterative procedure. Set v1=
v, V(Gl)={vi} ,
E(Gt) =0, S
1 ={vi} ,
and forifrom 2 to I
DI,find Vi in
D \ si-1with minimum
dc(vi, si-1)·Let
pibe
a shortest pathfrom Vi to
si-1; byLemma 2.13, this path
has length at most 3.At
each stepthe gra
phGi
is connectedb
causewe
are adding apath Pi which h
asone end
already inthe grap
h. TakeT to be a spanning
treeof t
he final graph GIDI·Note that
at eachstep
we add a vertex ofD
and atmost 3
edges toGi. Since there
are JDI-
1 iterations, this shows the graph CIDI (and consequently the tre T) has at most3(JDI -
1) edges.Note
also that V(T)=
V( GIDI) contains everyvertex of D. Hence
if we double the edges ofT we obtain aclos d dominating walk
of G of lengthat most 6(JD
J-1).Then
by Lemma 2.12 we know qJD
I guards can dominate
G leaving novertex
unobserved for more thanr
6(IDI - 1)1- 1 =
r~- _ 6
1 - 1::;r~q1 - 1
qJDI
qqJD
Iunits
of time, as claimed.
DLet us focus
now on
q = ~; i.e., supposethe
set of JDI guardshave
been cutby
half. We havethe following
result as an immediate corollary of Theorem 2.14.Corollary 2.15. If G is a connected graph with dominating set D then
l~ l
guards can monitor G such that no ver-tex is unobser-ved for- more than 11 units of time.We will see that 1~1 guards are ev n more effective if the vertices of D are suffi- ciently 'close' to one another; i.e., if we have a stronger condition than that guaranteed by Lemma 2.13. In this case we abandon the method of sharing a dominating walk.
The following theor ms explain how we can form clusters of the vertices of D and as- sign a number of guards to each, thereby reducing the total number of edges traversed (see Figure 2.6, for example). We need the following auxiliary graph.
Definition 2.16. Let G be a connected graph with dominating set D. For- a positive integer- d, define G D,d to be the gmph with ver-tex set D in which two ver-tices u, v E D ar-e adjacent if and only if de(u, v) :=:; d.
Theorem 2.17. [
6]
Let G be a connected gmph with dominating set D.(i) If de ( v, D \ {
v})
:=:; 2 for all v in D then 1 ~I guar-ds can monitor- G such that no vertex is unobser-ved for- mor-e than 7 units of time.(ii) If de(v, D \ {
v})
:=:; 1 for- all v in D then 1~1 guar-ds can monitor- G such that no ver-tex is unobserved for more than 3 units of time.Pmof. (i) Assume de( v, D \ { v}) :=:; 2 for all v in D; then by definition the graph G D,2
will have no isolates. Let M be a maximum matching in Gn,2 . All neighbours of an unmatched vertex are end vertices of an edge in Jl;f, since if two unmatched vertices are adjacent then their shared dge could belong to M, which contradicts the fact that M is maximum. For each unmatched vertex we can therefore select an edge incident with a matched neighbour. Now consid r an edge ·u, v of NI. If both u and v are incident with a selected edge then we have a path of length three, say P =
u' , u, v, v',
. - - - -- - - - - -- - - -~--- --- -
in
G
n,2 where u' and v' are not incident with edges ofNI .
ThenNI
could include the edges u'u and vv' instead of uv, again contradicting its maximality. Thus for each edge of M, exactly one end vertex is now connected to one or more unmatched vertices, thereby creating a collection of stars in G n,2 containing all vertic s of D. The edges in these stars represent paths of length at most 2 in G between two vertices of D.For each star on T vertices, double the edges on the corresponding paths in G and have l~J guards walk an Eulerian circuit in the resulting graph. There are T - 1 such paths, and when doubled each has length at most 4, so the guards follow each other along the circuit, spaced apart such that no vertex dominated by the walk is unobserved for more than
- 1< - 1= 7
r 4(T- lT/2J 1)1 - ~
(T-4 (T- 1)/ 1 ) 2 l
units of time. Since T is the number of vertices in each star of
G
n,2 and since these stars compris all vertices of D, placing l~J guards on each star in total uses at most1~1 guards.
( ii) If de (v, D \ { v}) ::; 1 for all v in D then G D ,1 has no isolates and we can form stars in this graph as described above. Each edge in a star corresponds to a single edge in G, so a star on T vertices shared by
l
~J
guards will have any two guards at mostJ2(T - 1) l < r 2(T - 1)
1 = 4
I lT/2J - (T - 1)/2
edges apart. Hence in this case l~l guards can monitor G such that no vertex is
unobserved for more than 3 units of time. 0
Note that this method of assigning guards to stars of the graph
Gn , d
can also beused if D satisfies only de ( v, D \ { v}) :::; 3 for all v in D (using graph G
n ,
3 ); however, in general there is no improvement in this case over the method of sharing a dominating walk. In particular, one finds only that no vertex is unobserved for mor than 11 units of time, which we already have from Corollary 2.15. However, the authors of [6]note that when many stars created in Theorem 2.17 have odd order r-, the number of guards used in total is actually significantly less than 1~1, since we reduce the number of guards on each odd star from r to l~J =
r2l
(recall that we initially assume every vertex of D has a guard, and that we downsize this set of guards by half). In these cases we can afford to 'waste' guards in certain parts of the graph, while still using only 1~1 in total. In particular, if in Theorem 2.17 we eliminate the condition that de(v, D \ { v}) :::; 2 or de(v, D \ { v}) :::; 1 or all v in D, then the resulting isolates in G n,2 or G D,l could be given their own guard provided there are at least as many odd stars as there are isolates. This gives the following corollary.Corollary 2.18. [6] Let G be a connected graph with dominating set D. Form a collection of stars in the graph G n,2 as descr-ibed in the pr-oof of Theorem 2.17; if the number- of odd stars is at least the number of isolates in G n,2 then
1~1
guards can monitor- G such that no ver-te:r; is unobserved for- more than 7 units of time. If de( v, D \ {v}) :::;
2 for all v E D and the number- of odd star-s in Gn,
1 is at least the number of isolates in G D,l then 1~1 guards can monitor- G such that no vertex is unobserved for more than 3 'units of time.Returning to Theorem 2.17, note that when D satisfies de(v, D \ { v}) :::; 1 for all v E D, D is a total dominating set. If the matching NJ defined in th proof of Th or m 2.17 is p rfect th n
D
is in fact a paired dominating set. The followingtheorem
shows how paired domination
is ideal for minimizing the length of time for which vertices are unobserved.Theorem 2.19.
[6]
A graph G can be monitored with -y~G) guards and leave no vertex unobserved for more than 1 unit of time if and only if G has a paired dominating set of sizery( G).
Proof. (
=>)
If Gcan
be monitored by 'Y~)guards such that every vertex is seen at
least onceevery two
units of time then theset S
1 of vertices occupid by the guards
at some timet and t
he set S2 of vertices occupied at timet +
1 must together form a dominating set of G;
i.e.,D = sl us2
is a dominating set ofG.
ThusID I 2:: ry(
G). Butsince there are -y~G) guards,
we must haveI S1I , IS2 I :::;
-y~G),so ID I :::;
-y(~)+
-y(~)=
ry(G)and consequently I DI =
ry(G). Weconclude that sl n s2 = 0, and if we
let Nf be theset
ofedges
walked by the guards,each
havingone end
vertex insl and one in s2 ,
then Nf is a perfect matching in the subgraph induced by D, and hence D is
a pair d
dominating set, as required.( {:::) If G has a paired dominating
set
Dthen
thesubgraph
induced by D hasa
perfect matching, M, whoseend
verticescomprise
D. Eachedg of
Nf can be traversed repeatedly by oneguard, so that
no vertex is unobserved for more than 1 unit of time,and this
method usesexactly
1~1guards. D
- - -- ----- - - -
Chapter 3
Fixed time
In this chapt r we explore a variation on the watchman's walk problem first introduced by Davies et al. in
[1].
Thisvariation takes t
he opposite standpoint of the problem discussed inChapter
2,assuming
that fixedtime constraints ar
imposed on the monitoring ofa graph G and attempting to determine t
he minimum number of guards, Wt (G), required to meet thoseconstraints.
We begin withan
introduction to this problem, including some basic results,
and proceed to findan upper bound on
Wt( G)for any odd integer t >
0.3.1 Introductory results
Recall that a graph G
can b
et-monitored by mguards if t
here exists a collection of mwalks (not
necessarily distinct ordisjoint) t
hat can betraversed by t
he guards sucht
hat no vertex in G is unobserved formore t
hant
units of time. Equivalently,every
vertex is either occupied by a guard oradj acent
toa
vertexoccupied
bya guard at
least once every t+
1 units of time.Suppose for example that the graph
G inFigure 3.1 below must be dominated such that no vertex is unobserved for more than t = 2 units of time. The gray vertices indicate the positions of four guards
91 ,9
2 , 93and
94at some fixed point in time, and t
hedotted arrows indicate the direction from which the guards have entered their current vertices. We will see
howt
hesefour guards can 2-monitor G. The guard 9
1traverses two edges, and all four of the vertices dominated by 9
1are seen at least once every 2 units of time. The guard 9
2 remains stationaryat the indicated vert x, thereby constantly dominating that vertex and its two neighbours. Guards
93and
94share a single closed walk, the intention being that the guards are spaced equally apart and follow one another along the walk. The reader can v rify that this ensures no vertex dominated by
93and
94 isunobserved for more than t = 2 units of time.
G
Figure 3.1: A graph
Gthat is 2-monitored with four guards.
In Figure
3.1, each guard repeatedly traverses a
closed walk. Althoughthere is
no such stipulation in the
definitionof t-monitoring, we may in fact assume this is
always the case.
Let Gbe a graph t-monitored by guards and suppose one or more
of these guards share a walk W that is not closed. At any fixed point in time, label a vertex 0* if it is currently occupied by a guard, label a vertex 0 if it is unoccupied but adjacent to a vertex with a guard, and label every other vertex with a positive integer (at most t) according to the length of time since the vertex was last observed. For example, from the graph G in Figure 3.1 we obtain the vertex labelling shown in Figure 3.2 below. Since both t and
IV(T)I
are finite, there are only finitely many such labellings, and so at some point a vertex labelling will be repeated. When this happens, we can truncateW
and have it repeat whatev r edge sequence followed the first occurrence of that labelling. The new walk is closed does not disrupt the t-monitoring ofG.
Since any non-closed walk can be reconstructed in thi way, have the following theorem.G
0 1 0
o"
Figure 3.2: The length of time for which each vertex in G has been unobserved.
Theorem 3.1. If a graph G can bet-monitored by m guards then G can bet-monitored by m guards whose walks are closed.
ote that for any connected graph G, if m guards can (minimally) monitor G such t hat each vertex
is seen within everyt + 1 units of t ime, t hen with those
m guardseach vertex
is also seen within every t + 2 units of time. Thus for any t , we have
Wt (G) ~ Wt+ 1(G). This
idea issummari zed
inLemma 3. 2 and will be useful as w
investigat e increasing values oft.
Lemma 3.2.
For any graph G
7 T!V0(G) ~VV
1(G )
~ W2(G) ~ ....In [1], the a
uthors discuss bounds on W
t (G) for various values of t, wit h G usually assum d t o b e a tree. vVe b egin naturally with t = 0;
inthis case,
Wt (G)= I'( G) , since if vertices cannot b e
unobserved foreven a single unit of time then the guards must dominate all vertices while remaining stationary.
In this section , when a graph
G isclear from the context , let
n representI V (G)
1.Theorem 3.3.
[1] For any connected graph G
7W
0(G )::::;
l~J.Proo f. For any dominating set
Dof a conn ct d gr aph G , the set
V(G)\Dis also a dominating set of G.
Hence aminimum dominating set must have cardinality
lesst han or equ al t o
l~J,as otherwise its se t complement is a dominating set wit h fewer vertices . Thus I' ( G)::::;
l~J,and since
W0(G)=I'( G) , the result foll ows. 0
The trees of even order
nthat have domination number equal to
~are classified
in [2]: they are composed ofan equal number of
leaves and non-leaves,wi t h every non-leaf adjacent to exactly one
leaf. Becauseof t
he complexityof t he t ime restraint problem for gen eral graphs, the remainder of this chapter predominantly considers trees.
Theorem 3.4 summarizes three importa